Equation of a plane orthogonal to a vector

[starbaj12]

yes I agree, and I tried to go further to see if I could work something out but still no lightbulb yet

 

[arildno]

Ok, so rewriting (eq.2) a bit, we have:
$$\vec{W}=\vec{F}+\vec{B}$$
(You've seen this one before..)

We have already established, that all acceptable choices of $$\vec{F}$$ together constitutes a plane, with $$\vec{B}$$ as the normal vector.

Now then, if you combine this insight with the above equation, what sort of geometric surface must all acceptable choices for $$\vec{W}$$ taken together constitute?

 

[starbaj12]

A plane i think

 

[starbaj12]

a plane parellel to the normal vector b

 

[arildno]

It sure is!
Let us argue like this:
1) Take any acceptable choice of $$\vec{F}$$
(Such a point lies in the plane containing the origin, with $$\vec{B}$$ as normal vector.
2) To find the corresponding $$\vec{W}$$ we go straight up along $$\vec{B}$$
(Adding that is, $$\vec{B}$$ to our $$\vec{F}$$)
3). Now, the DISTANCE that $$\vec{W}$$ is removed from the plane in which $$\vec{F}$$ lies, is how far along the vector normal to the plane $$\vec{W}$$ is.
4) But this distance is simply the magnitude of $$\vec{B}$$, since $$\vec{B}$$ IS normal to the plane in which $$\vec{F}$$ lies.
5) Clearly, therefore, EVERY ACCEPTABLE CHOICE OF $$\vec{W}$$ LIES AN EQUAL DISTANCE FROM OUR PLANE (that is given by the length of $$\vec{B}$$
6) But this means, that the set of acceptable $$\vec{W}$$ is A PLANE PARALLELL TO THE PLANE OF ACCEPTABLE $$\vec{F}$$!

Do you agree with this reasoning?

 

[starbaj12]

I'm not understanding #5

 

[arildno]

1.Each acceptable choice of $$\vec{W}$$ corresponds to an acceptable choice of $$\vec{F}$$
2. The distance of a point to a plane is found by measuring the length of the perpendicular (normal) joining that point to the plane.
3. In our case, the perpendicular to the plane in which our $$\vec{F}$$ lies, is given by $$\vec{B}$$

Do you agree with this?

 

[starbaj12]

yes I'm following now

 

[arildno]

So, you accept that the set of acceptable $$\vec{W}$$ represents a plane with normal vector given by $$\vec{B}$$ ?

 

[starbaj12]

yes, but this is due in another twenty minutes if you do not mind could you walk me through the steps. Another thing this is a physics assingment, but I placed it in the math due to it being math oriented. I took intro to linear algebra and I realize that in linear algebra it has to be more in dept, but for physics not as much. But I really appreciate your help, nothing like understanding a problem fully through.

 

[arildno]

But now we're really done!
Because:
1) We've shown that the set of acceptable $$\vec{W}$$ is a plane.
2) In addition, since $$\vec{F}=\vec{0}$$ was a solution of (eq.3), we know that one acceptable $$\vec{W}=\vec{0}+\vec{B}=\vec{B}$$
But the physical point corresponding to $$\vec{B}$$ is D..

This was what you had to show:
That your original equation represents a plane with normal vector given by $$\vec{B}$$ containing D.

 

[starbaj12]

I had to leave to fast I just want to say thank you for all your help arildno